AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Applications of first law of thermodynamics.
Full specification wording
First law of thermodynamics (A-level only): Applications of first law of thermodynamics.
Apply Q = ∆U + W to real processes.
Checked against the AQA 7408 specification (opens in a new tab) on .
- Point 2 of 2
- 2 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Apply Q = ∆U + W to real processes. In an isothermal expansion of an ideal gas, ∆U = 0, so Q = W: heating supplies the energy for the gas to do work. In adiabatic compression, Q = 0, so ∆U = −W; work done on the gas increases internal energy and temperature. In a rigid container, W = 0, so Q = ∆U. In free expansion into a vacuum, W = 0 and Q = 0, so ∆U = 0. For a cyclic process, ∆U = 0 over a full cycle, so net Q = net W. Use these cases to deduce energy transfers, temperature changes and work done, and to explain everyday examples such as a bicycle pump warming up or a gas cylinder cooling as it discharges.
What earns the marks
- Applies Q = ∆U + W to an isothermal process, stating ∆U = 0 and hence Q = W.
- Applies Q = ∆U + W to an adiabatic process, stating Q = 0 and hence ∆U = −W, linking work done on a gas to a temperature rise.
- Applies Q = ∆U + W to a constant-volume (isochoric) process, stating W = 0 and hence Q = ∆U.
- Explains a cyclic process by stating that ∆U = 0 over a complete cycle, so the net heating equals the net work done by the system.
- Uses the first law to explain a practical example, such as a bicycle pump warming when air is compressed quickly or a gas canister cooling as it discharges.
Common mistakes and how to put them right
Assuming Q = 0 for any rapid process.
Put it right: Check whether the process is thermally insulated or whether heating has time to occur.
Forgetting that ∆U = 0 for an isothermal process only for an ideal gas.
Put it right: State the ideal-gas assumption.
Confusing constant-volume with constant-pressure processes.
Put it right: Use W = 0 only when volume is constant.
Exam tips
- Identify the process type first (isothermal, adiabatic, isochoric, cyclic) and write the resulting simplification before calculating.
- For adiabatic processes, state clearly that Q = 0 and link the sign of W to whether the gas is compressed or expands.
- When explaining a practical example, name the process and justify the simplification using the first law.
Try the lesson
The start of the First law of thermodynamics (A-level only): Applications of first law of thermodynamics. lesson, made from checked teaching notes for this specification point.
Have a go first
A gas is heated in a rigid sealed container so that its volume cannot change. Energy is transferred to the gas by heating only. Applying the first law of thermodynamics, which quantity is zero?
Listen
Apply Q = ∆U + W to real processes. In an isothermal expansion of an ideal gas, ∆U = 0, so Q = W: heating supplies the energy for the gas to do work. In adiabatic compression, Q = 0, so ∆U = −W; work done on the gas increases internal energy and temperature. In a rigid container, W = 0, so Q = ∆U. In free expansion into a vacuum, W = 0 and Q = 0, so ∆U = 0. For a cyclic process, ∆U = 0 over a full cycle, so net Q = net W. Use these cases to deduce energy transfers, temperature changes and work done, and to explain everyday examples such as a bicycle pump warming up or a gas cylinder cooling as it discharges. A strong answer earns marks for these points. Applies Q = ∆U + W to an isothermal process, stating ∆U = 0 and hence Q = W. Applies Q = ∆U + W to an adiabatic process, stating Q = 0 and hence ∆U = −W, linking work done on a gas to a temperature rise. Applies Q = ∆U + W to a constant-volume (isochoric) process, stating W = 0 and hence Q = ∆U. Explains a cyclic process by stating that ∆U = 0 over a complete cycle, so the net heating equals the net work done by the system. A common slip: Assuming Q = 0 for any rapid process: the error is treating a fast process as automatically adiabatic. Instead, check whether the process is thermally insulated or whether heating has time to occur. Exam tip: Identify the process type first (isothermal, adiabatic, isochoric, cyclic) and write the resulting simplification before calculating.
Key words
Say what each word means before you turn the card.
Isochoric process
Try answering aloud or in your head before turning the card.
Cyclic process
Try answering aloud or in your head before turning the card.
See every definition
- Isochoric process
- A constant-volume process in which W = 0, so Q = ∆U.
- Cyclic process
- A process in which ∆U = 0 over a full cycle, so net Q = net W.
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